Could you explain how to show 10.4 in great detail? I also included a list of theorems and definitions in my textbook as a reference. Really appreciate your help! Definition 10.1. Let R be a ring. Then a polynomial with coefficients in R is aformal expressionao + ajx + a2x² + · · · + anx",where a; e Rand n is a nonnegative integer. Suppose that bo +bjx + • ·.+ bmx™is also a polynomial with coefficients in R. Without loss of generality, let us say thatn < m. Then these polynomials are equal if and only if a; = b; for all i n. The set of all polynomials with coefficients in R is denotedR[x].Example 10.1. Let R = Z5. Then (inserting congruence class brackets for clarity),an example of a polynomial in R[x] would be f (x) = [3] + [2]x + [4]x². As partof the above definition, we observe that f (x) = g(x), where g(x) = [3] + [2]x +[4]x? + [0]x³.= ao + a¡x + · · · + a,,x" e R[x].Definition 10.2. Let R be a ring and let f(x)Further suppose that am +0 but ag = 0 for all k > m. Then the degree of f (x) is m,and we write deg(f (x)) = m. The leading term of f (x) is a,mx", and the leadingcoefficient is a,m. Note that the zero polynomial, 0, has no degree, leading term orleading coefficient. A constant polynomial has degree 0 (or is the zero polynomial).If R has an identity, then f (x) is monic if its leading coefficient is 1.= 3+ 7x – 15x? + 0x³ + 2x4 + 0x°. ThenExample 10.2. In Q[x], let ƒ(x)deg(f (x)) = 4, the leading term is 2x* and the leading coefficient is 2. This poly--nomial is not monic.We wish to make R[x] into a ring, and so we need addition and multiplicationoperations. These will be exactly the same as for real polynomials. Let f(x) =ao + ajx + · ..+ a„x" and g(x) = bo + bịx + ...+ bmx". Adding in terms withzero coefficients if necessary, we may assume that m= n. Thenf (x) + g(x) = (ao + bo) + (ai + bj)x + · . - + (a, + b„)x".Similarly,m+nf (x)g(x) = co + c1x + c2x² +… .+ cm+nx'whereC; = aob; + ab¡-1+a2b¡-2 +…+a;bo.Here, we take a ; = 0 if j > n and b; = 0 if j > m.= 5+2x+6x² and g (x) = 3+x+4x²+5x³. ThenExample 10.3. In Z7[x], let ƒ (x)f (x)+g(x) = 1+3x+3x²+5x³ and ƒ (x)g(x) = 1+4x+5x² +4x³+6x4+2xS.Theorem 10.1. If R is a ring, then so is R[x].Corollary 10.1. Let R be a ring. Then1. if R has an identity, then so does R[x]; and2. if R is commutative, then so is R[x].Theorem 10.2. Let R be an integral domain, and let f(x) and g(x) be nonzeropolynomials in R[x], of degree m and n respectively. Then1. deg(f (x) + g(x)) is at most the larger of m and n (or f (x)+g(x) = 0); and2. deg(ƒ(x)g(x)) :— т + п.Corollary 10.2. If R is an integral domain, then so is R[x].Theorem 10.3. (Division Algorithm for Polynomials). Let F be a field, and letf (x), g(x) E F[x], with g(x) # 0. Then there exist unique q(x), r(x) E F[x] suchthatf (x) = g(x)q(x)+r(x),with either r(x) = 0 or deg(r(x)) < deg(g(x)). 10.4. Let R be an integral domain. Show that the units of R[x] are precisely theconstant polynomials a, where a e U(R).

Let R, +, · be a ring with unity and a∈R. Then, a is said to be unit element of R if there exists b∈R such that a·b=b·a=1. In other words , a non zero element of a ring which has multiplicative inverse is called a unit. The set of all units of a ring R is denoted by UR. A commutative ring R with unity is called an integral domain if a·b=0 ⇒ a=0 or b=0 for a,b∈R. Let R be a ring then the set Rx=a0+a1x+a2x2+. . . +anxn : ai∈R and n is a non negative integer forms a ring with respect to addition and multiplication of polynomials. If R is an integral domain, then the polynomial ring Rx is also an integral domain and a unit in R is a unit in Rx Note that a constant polynomial has a degree 0. Given that R is an integral domain. It is required to show that the units of Rx are precisely the constant polynomials a, where a∈UR Let fx be a unit in Rx. Then there exist a polynomial gx∈Rx such that f(x)·gx=1 ............(1) Let the degree of fx and gx be m and n respectively. Then by theorem 10.2, degfxgx=m+n The degree of the constant polynomial on R.H.S is 0 i.e., deg1=0 From (1) , f(x)·gx=1 Then degfxgx=deg1 ⇒ m + n = 0 Since the degree is a non negative function, the only possibility are m=n=0 Then the degree of f(x) and g(x) is zero i.e., f(x) and g(x) are constant polynomials. Let f(x)=a and g(x)=b with a,b ∈R and a.b=1 This implies that a is a unit in R i.e., a∈UR Thus the unit fx of Rx is the constant polynomial a, where a∈UR Hence it is proved that the units of Rx are precisely the constant polynomials a, where a∈UR